Given a triangle ABC, A (-5; -2) B (-2; 2) C (3; 13) find cos A.

Let a triangle ABC be given, the vertices of which have coordinates A (- 5; – 2); B (- 2; 2); C (3; 13), then:

(- 2 – (- 5); 2 – (- 2)) = (3; 4) – coordinates of the vector AB;

(3 – (- 5); 13 – (- 2)) = (8; 15) – coordinates of the AC vector;

√ (3² + 4²) = 5 – the module of the vector AB;

√ (8² + 15²) = 17 – the module of the vector AC;

3 ∙ 8 + 4 ∙ 15 = 84 – scalar product of vectors AB and AC;

cos А = 84 / (5 ∙ 17) = 84/85, since the cosine of the angle between the vectors is equal to the ratio of the scalar product of vectors to the product of the moduli of these vectors.

Answer: cos A = 84/85.